Problem: $\int x\sqrt{5+2x^2}\, \,dx\,=$ $+~C$
Explanation: Notice that we can rewrite the integral as $\int\sqrt{5+2x^2}\,\cdot x \,dx\,$. If we let $ {u=5+2x^2}$, then ${du=4x\, dx}$ and $x\,dx=\dfrac{du}{4}}$. Substituting gives us: $\begin{aligned}\int\sqrt{{5+2x^2}}\,\cdot x \,dx}\,&= \int\sqrt{ u}\,\,\cdot \dfrac{du}{4}}\,\\\\\\ &= \int\sqrt{ u} \cdot \dfrac14\, du \\\\\\ &=\dfrac14 \int\sqrt{ u}\, du\end{aligned}$ We recognize this antiderivative. ∫ e x 1 + e 2 x d x = 1 4 ∫ u √ d u = 1 4 ⋅ 2 3 u 3 2 + C = 1 6 u 3 2 + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \dfrac14 \int\sqrt{ u}\, du\\\\\\ &=\dfrac14\cdot \dfrac23u\^{\frac32}+C\\\\\\ &=\dfrac{1}{6}u\^{\frac32}+C\end{aligned} We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = 1 6 u 3 2 + C = 1 6 ( 5 + 2 x 2 ) 3 2 + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac{1}{6}u\^{\frac32}+C\\\\\\\ &=\dfrac{1}{6}(5+2x^2)\^{\frac32}+C\end{aligned} The answer: ∫ x 5 + 2 x 2 − − − − − − √ d x = 1 6 ( 5 + 2 x 2 ) 3 2 + C \int x\sqrt{5+2x^2}\, \,dx\,=\dfrac{1}{6}(5+2x^2)\^{\frac32}+C